What is the integral of ((e^{-x})^2)/b ?

The integral of ((e^{-x})^2)/b is -(e^{-2x})/(2b)+C

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integral of ((e^{-x})^2)/b

integral

\frac{( e ^{- x} ) ^{2}}{b}

- \frac{e ^{- 2 x}}{2 b} + C

Solution steps

\int \frac{( e ^{- x} ) ^{2}}{b} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{b} \cdot \int (e^{- x})^{2} d x

Apply u-substitution

: \int - u d u

= \frac{1}{b} \cdot \int - u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{b} (- \int u d u)

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{b} (- \frac{u ^{2}}{2})

Substitute back

u = e^{- x}

= \frac{1}{b} (- \frac{( e ^{- x} ) ^{2}}{2})

Simplify

\frac{1}{b} (- \frac{( e ^{- x} ) ^{2}}{2}) : - \frac{e ^{- 2 x}}{2 b}

= - \frac{e ^{- 2 x}}{2 b}

Add a constant to the solution

= - \frac{e ^{- 2 x}}{2 b} + C

5 t^{2} - t

\frac{1}{ln ^{2} ( x )}

x csc^{5} (x^{2})

sec^{2} (2 x) - 1

\frac{t ^{3}}{t ^{2} + 1}

What is the integral of ((e^{-x})^2)/b ?
The integral of ((e^{-x})^2)/b is -(e^{-2x})/(2b)+C